1. 7 -1 Chapter 7 Dynamic Programming

2. 7 -2 Fibonacci sequence (1) 0,1,1,2,3,5,8,13,21,34,... Leonardo Fibonacci (1170 -1250) 用來計算兔子的數量 每對每個月可以生產一對 兔子出生後, 隔一個月才會生產, 且永不死亡 生產 0 1 1 2 3... 總數 1 1 2 3 5 8... http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibnat.html

3. 7 -3 Fibonacci sequence (2) 0,1,1,2,3,5,8,13,21,34,...

4. 7 -4 Fibonacci sequence and golden number 0,1,1,2,3,5,8,13,21,34,... f n = 0 if n = 0 f n = 1 if n = 1 f n = f n-1 + f n-2 if n  2 1 x-1 x

5. 7 -5 Computation of Fibonacci sequence Solved by a recursive program: Much replicated computation is done. It should be solved by a simple loop. f n = 0 if n = 0 f n = 1 if n = 1 f n = f n-1 + f n-2 if n  2

6. 7 -6 Dynamic Programming Dynamic Programming is an algorithm design method that can be used when the solution to a problem may be viewed as the result of a sequence of decisions

7. 7 -7 The shortest path To find a shortest path in a multi-stage graph Apply the greedy method : the shortest path from S to T : 1 + 2 + 5 = 8

8. 7 -8 The shortest path in multistage graphs e.g. The greedy method can not be applied to this case: (S, A, D, T) 1+4+18 = 23. The real shortest path is: (S, C, F, T) 5+2+2 = 9.

9. 7 -9 Dynamic programming approach Dynamic programming approach (forward approach): d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} d(A,T) = min{4+d(D,T), 11+d(E,T)} = min{4+18, 11+13} = 22.

10. 7 -10 d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F, T)} = min{9+18, 5+13, 16+2} = 18. d(C, T) = min{ 2+d(F, T) } = 2+2 = 4 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} = min{1+22, 2+18, 5+4} = 9. The above way of reasoning is called backward reasoning.

11. 7 -11 Backward approach (forward reasoning) d(S, A) = 1 d(S, B) = 2 d(S, C) = 5 d(S,D)=min{d(S,A)+d(A,D), d(S,B)+d(B,D)} = min{ 1+4, 2+9 } = 5 d(S,E)=min{d(S,A)+d(A,E), d(S,B)+d(B,E)} = min{ 1+11, 2+5 } = 7 d(S,F)=min{d(S,B)+d(B,F), d(S,C)+d(C,F)} = min{ 2+16, 5+2 } = 7

12. 7 -12 d(S,T) = min{d(S, D)+d(D, T), d(S,E)+ d(E,T), d(S, F)+d(F, T)} = min{ 5+18, 7+13, 7+2 } = 9

13. 7 -13 Principle of optimality Principle of optimality: Suppose that in solving a problem, we have to make a sequence of decisions D 1, D 2, …, D n. If this sequence is optimal, then the last k decisions, 1  k  n must be optimal. e.g. the shortest path problem If i, i 1, i 2, …, j is a shortest path from i to j, then i 1, i 2, …, j must be a shortest path from i 1 to j In summary, if a problem can be described by a multistage graph, then it can be solved by dynamic programming.

14. 7 -14 Forward approach and backward approach: Note that if the recurrence relations are formulated using the forward approach then the relations are solved backwards. i.e., beginning with the last decision On the other hand if the relations are formulated using the backward approach, they are solved forwards. To solve a problem by using dynamic programming: Find out the recurrence relations. Represent the problem by a multistage graph. Dynamic programming

15. 7 -15 The resource allocation problem m resources, n projects profit P i, j : j resources are allocated to project i. maximize the total profit.

16. 7 -16 The multistage graph solution The resource allocation problem can be described as a multistage graph. (i, j) : i resources allocated to projects 1, 2, …, j e.g. node H=(3, 2) : 3 resources allocated to projects 1, 2.

17. 7 -17 Find the longest path from S to T : (S, C, H, L, T), 8+5+0+0=13 2 resources allocated to project 1. 1 resource allocated to project 2. 0 resource allocated to projects 3, 4.

18. 7 -18 The longest common subsequence (LCS) problem A string : A = b a c a d A subsequence of A: deleting 0 or more symbols from A (not necessarily consecutive). e.g. ad, ac, bac, acad, bacad, bcd. Common subsequences of A = b a c a d and B = a c c b a d c b : ad, ac, bac, acad. The longest common subsequence (LCS) of A and B: a c a d.

19. 7 -19 The LCS algorithm Let A = a 1 a 2  a m and B = b 1 b 2  b n Let L i,j denote the length of the longest common subsequence of a 1 a 2  a i and b 1 b 2  b j. L i,j = L i-1,j-1 + 1 if a i =b j max{ L i-1,j, L i,j-1 } if a i  b j L 0,0 = L 0,j = L i,0 = 0 for 1  i  m, 1  j  n.

20. 7 -20 The dynamic programming approach for solving the LCS problem: Time complexity: O(mn)

21. 7 -21 Tracing back in the LCS algorithm e.g. A = b a c a d, B = a c c b a d c b After all L i,j ’ s have been found, we can trace back to find the longest common subsequence of A and B.

22. 7 -22 0/1 knapsack problem n objects, weight W 1, W 2, ,W n profit P 1, P 2, ,P n capacity M maximize subject to  M x i = 0 or 1, 1  i  n e. g.

23. 7 -23 The multistage graph solution The 0/1 knapsack problem can be described by a multistage graph.

24. 7 -24 The dynamic programming approach The longest path represents the optimal solution: x 1 =0, x 2 =1, x 3 =1 = 20+30 = 50 Let f i (Q) be the value of an optimal solution to objects 1,2,3, …,i with capacity Q. f i (Q) = max{ f i-1 (Q), f i-1 (Q-W i )+P i } The optimal solution is f n (M).

25. 7 -25 Optimal binary search trees e.g. binary search trees for 3, 7, 9, 12;

26. 7 -26 Optimal binary search trees n identifiers : a 1 <a 2 <a 3 < … < a n P i, 1  i  n : the probability that a i is searched. Q i, 0  i  n : the probability that x is searched where a i < x < a i+1 (a 0 =- , a n+1 =  ).

27. 7 -27 Identifiers : 4, 5, 8, 10, 11, 12, 14 Internal node : successful search, P i External node : unsuccessful search, Q i The expected cost of a binary tree: The level of the root : 1

28. 7 -28 The dynamic programming approach Let C(i, j) denote the cost of an optimal binary search tree containing a i, …,a j. The cost of the optimal binary search tree with a k as its root :

29. 7 -29 General formula

30. 7 -30 Computation relationships of subtrees e.g. n=4 Time complexity : O(n 3 ) (n-m) C(i, j) ’ s are computed when j-i=m. Each C(i, j) with j-i=m can be computed in O(m) time.

31. 7 -31 Matrix-chain multiplication n matrices A 1, A 2, …, A n with size p 0  p 1, p 1  p 2, p 2  p 3, …, p n-1  p n To determine the multiplication order such that # of scalar multiplications is minimized. To compute A i  A i+1, we need p i-1 p i p i+1 scalar multiplications. e.g. n=4, A 1 : 3  5, A 2 : 5  4, A 3 : 4  2, A 4 : 2  5 ((A 1  A 2 )  A 3 )  A 4, # of scalar multiplications: 3 * 5 * 4 + 3 * 4 * 2 + 3 * 2 * 5 = 114 (A 1  (A 2  A 3 ))  A 4, # of scalar multiplications: 3 * 5 * 2 + 5 * 4 * 2 + 3 * 2 * 5 = 100 (A 1  A 2 )  (A 3  A 4 ), # of scalar multiplications: 3 * 5 * 4 + 3 * 4 * 5 + 4 * 2 * 5 = 160

32. 7 -32 Let m(i, j) denote the minimum cost for computing A i  A i+1  …  A j Computation sequence : Time complexity : O(n 3 )

33. 7 -33 Single step graph edge searching fugitive: can move in any speed and is hidden in some edge of an undirected graph G=(V,E) edge searcher(guard): search an edge (u, v) from u to v, or stay at vertex u to prevent the fugitive passing through u Goal: to capture the fugitive in one step. no extra guards needed extra guards needed

34. 7 -34 cost of a searcher from u to v: wt(u) a guard staying at u: wt(u) Cost of the following: 2wt(a)+wt(b)+wt(b) (one extra guard stays at b) Problem definition: To arrange the searchers with the minimal cost to capture the fugitive in one step. NP-hard for general graphs; P for trees.

35. 7 -35 The weighted single step graph edge searching problem on trees T(v i ): the tree includes v i, v j (parent of v i ) and all descendant nodes of v i. C(T(v i ), v i, v j ): cost of an optimal searching plan with searching from v i to v j. C(T(v 4 ), v 4, v 2 )=5 C(T(v 4 ), v 2, v 4 )=2 C(T(v 2 ), v 2, v 1 )=6 C(T(v 2 ), v 1, v 2 )=9

36. 7 -36 The dynamic programming approach Rule 1: optimal total cost Rule 2.1 : no extra guard at root r: All children must have the same searching direction.

37. 7 -37 Rule 2.2: one extra guard at root r: Each child can choose its best direction independently.

38. 7 -38 Rule 3.1 : Searching to an internal node u from its parent node w

39. 7 -39 Rule 3.2 : Searching from an internal node u to its parent node w

40. 7 -40 Rule 4: A leaf node u and its parent node w. the dynamic programming approach: working from the leaf nodes and gradually towards the root Time complexity : O(n) computing minimal cost of each sub-tree and determining searching directions for each edge